Copyright (c) Bani K. Mallick1 STAT 651 Lecture #17.

Copyright (c) Bani K. Mallick 1

STAT 651

Lecture #17


Topics in Lecture #17 Chi-squared tests for independence


Book Sections Covered in Lecture #17

Chapter 10.6


Lecture 17 Review: Comparison of Two Population Proportions

In some cases, we may want to compare two populations 1 and 2

The null hypothesis is H0: 1 = 2

This is the same as H0: 1 - 2 = 0



The null hypothesis is H0: 1 - 2 = 0

Form a CI for the difference in population proportions 1 - 2

The estimate of this difference is simply the difference in the sample fractions:

1 2ˆ ˆ



The estimated standard error of the difference in the sample fractions:

The (1100% CI then is

2

1 1 2 2

1 2

1 1

1ˆ ˆ

( ) ( )ˆ ˆ ˆ ˆˆ

n n

21 2 2 1/ ˆ ˆzˆ ˆ ˆ


Lecture 17 Review: Comparison of Two Population Proportions:

Remarkably, but perhaps not surprisingly, you do not have to compute these confidence intervals by hand!

The idea: simply pretend, and I do mean pretend, that the binary outcomes are real numbers and run your ordinary t-test CI, unequal variance line


Chisquared Tests for Independence and Homogeneity

In the previous lecture, we asked whether two populations has the same fraction (proportion)

Thus, the populations were (1) very good beers (2) good or fair beers

We looked at the proportion of beer that were widely available in the U.S.



Thus, the populations were (1) very good beers (2) good or fair beers

We looked at the proportion of beers that were widely available in the U.S.

If the proportions of the population that are widely available are the same, then the proportion of beer that is widely available is independent of whether the beer of very good or just good/fair.



If the proportions of the population that are widely available are the same, then the proportion of beer that is widely available is independent of whether the beer of very good or just good/fair.

Thus, we can think about testing whether the outcomes (widely available or not) are independent of the populations (very good or fair/good)



We can test whether two categorical factors (availability and beer rating) are independent or not

The null hypothesis is that they are independent

The alternative hypothesis is that they are not independent

This can be tested using a chisquared test

0H

aH



The chisquared test for independence of two categorical factors:

The factors can have more than 2 levels: this is the main advantage of the chisquared test

Thus, for example, you could define three populations of beers (fair, good, very good), three levels of availability (special, regional, national) and ask whether quality is independent of availability.



In SPSS, you get a table with counts in each “cell” and along the rows and columns

# of observations in row i, column j =

# of observations in row i =

# of observations in column j =

# of observations =

ijn

in

jn

n


Factor A

Level 1 Level 2

Level 1Factor B

Level 2

11n

This describes the populationtable in its most general formin terms of counts

12n

21n 22n

11 12 1n n n

21 22 2n n n

11 21 1n n n 12 22 2n n nn


US Availability and Rating: Are Better Beers More Widely

Available?Availability in the U.S. * Very Good versus Other Crosstabulation

Count

6 6 12

5 18 23

11 24 35

National

Regional

Availabilityin the U.S.

Total

Very Good Fair or Good

Very Good versus Other

Total

The are 6 Very Good, National beers in this sample.

There are 11 Very Good beers in the sample

The total sample is of size 35



For purposes of explanation, let’s make up a fake example.

Consider two categorical factors for males: height (short, tall) and favorite sport (golf, baseball)

We probably would not expect these to be related.

Here is the data table (next slide!)


Sport Preference

Golf Baseball

ShortHeight

Tall

This describes the population table in its most general form in terms of row and column counts only.

There are 400 short men, 200 golfers, etc.

1000

400

600

200 800


Sport Preference

Golf Baseball

ShortHeight

Tall

Under the null hypothesis that height and sport preference areindependent, how many short men who play golf would youExpect? Think hard about this!

1000

400

600

200 800


Sport Preference

Golf Baseball

ShortHeight

Tall

80

Note that 40% of men are short. Under the null hypothesis that height and sport preference are independent, of the 200 men who prefer golf, you would expect 40% = 80 to be short.

1000

400

600

200 800

Expected Cell Counts under the null hypothesis


Sport Preference

Golf Baseball

ShortHeight

Tall

80

Note that 40% of men are short. Under the null hypothesis that height and sport preference are independent, of the 800 men who prefer baseball, you would also expect 40% = 320 to be short.

1000

400

600

200 800

320



Sport Preference

Golf Baseball

ShortHeight

Tall

80

Under the null hypothesis that height and sport preference are independent, you can fill out the rest of the table ofexpected counts

1000

400

600

200 800

320


120 480


Sport Preference

Golf Baseball

ShortHeight

Tall

80 & 100

Now you have to ask yourself, are the observed counts and the expected counts (under independenceunder independence) sufficiently different as to make the null hypothesis very unlikely?

1000

400

600

200 800

320 & 300


and the Observed Counts

120 & 100

480 & 500



The expected number of observations in any cell of the table is

You simply multiply the row and column totals and divide by the total sample size.

The chisquared test for independence and homogeneity simply compares the actual table fractions/numbers (observed) to the table fractions/numbers you would expect (expected) under the null hypothesis of independence

i jn n / n


Sport Preference

Golf Baseball

ShortHeight

Tall

80 & 100

Note how the expected count for the number of men who are tall and prefer baseball is 600 x 800 / 1000 = 480

1000

400

600

200 800

320 & 300


and the Observed Counts

120 & 100

480 & 500



The test statistic is computed as follows

First get the expected counts under independence

Compute 2

2

sum over all cells in the table

(Observed - Expected)Expected



The test statistic is 2

2

i j

ij

i jrows i,columns j

n nn

nn n

n

This has the counts in the cells and their expected values under the null hypothesis



If the Table has r-rows and c-columns:

The test statistic is

You reject the null hypothesis at Type I error (level) if >

Here “cuts off” area in the chisquared distribution with (r-1)x(c-1) degrees of freedom (Table 7)

2

2 21 1 ,(r ) (c )

21 1 ,(r ) (c )



The chisquared statistic can be computed in SPSS, by going to “analyze”, “descriptives” “crosstabs”

Then click on “statistics” and ask for “chisquared”

The p-value is slightly different from the p-value using the t-test method, but is generally pretty close. If in dispute, use the Pearson chisquared reading, with 1 exception



SPSS will print out a message if the expected count is < 5 in any cell, i.e.,

In this case, use the Fisher exact value

If Fisher’s exact value does not exist in a package, use the likelihood ratio test p-value

5 i jn n / n




Count

6 6 12

5 18 23

11 24 35

National

Regional


Total



Total

This is the table of observed counts. Under the null hypothesis that availability and beer quality are independent, how many very good, national beers would you expect?




Count

6 6 12

5 18 23

11 24 35

National

Regional


Total



Total

This is the table of observed counts. Under the null hypothesis that availability and beer quality are independent, how many very good, national beers would you expect?

12 x 11 / 35 = 3.77




Count

6 & 3.77 6 & 8.23 125 & 7.23 18 & 15.77 23

11 24 35

National

Regional


Total



Total

This is the table of observed counts, with the expected counts under the null hypothesis that availability and beer quality are independent




Count

6 & 3.77 6 & 8.23 125 & 7.23 18 & 15.77 23

11 24 35

National

Regional


Total



Total

The chisquared statistic is

( (6-3.77) x (6-3.77)/3.77)

+ ( (6 – 8.23) x (6-8.23) / 8.23) + ((5-7.23) x (5-7.23)/7.23)

+ ((18-15.77) x (18-15.77)/15.77) = 2.9




Count

6 & 3.77 6 & 8.23 125 & 7.23 18 & 15.77 23

11 24 35

National

Regional


Total



Total

The chisquared statistic is = 2.9.

Here r = 2, c = 2, (r-1) x (c-1) = 1, and the critical value from Table 7 is 3.8416.

Note though that an expected cell count is < 5, so you have to use

Fisher’s exact value



Available?: p = 0.130 Note the warning message

Chi-Square Tests

2.922 b 1 .087

1.758 1 .185

2.854 1 .091

.130 .094

2.839 1 .092

35

Pearson Chi-Square

Continuity Correction a

Likelihood Ratio

Fisher's Exact Test

Linear-by-LinearAssociation

N of Valid Cases

Value dfAsymp. Sig.(2-sided)

Exact Sig.(2-sided)

Exact Sig.(1-sided)

Computed only for a 2x2 tablea.

1 cells (25.0%) have expected count less than 5. The minimum expected count is3.77.

b.

Note the warning message in red, indicating the need to use Fisher’s exact test



SPSS also gives you expected counts and percentages in the table.

You ask for “Cells” and then click on what you want

SPSS demo


Chisquared Tests for Independence

Availability in the U.S. * Very Good versus Other Crosstabulation

6 6 12

3.8 8.2 12.0

50.0% 50.0% 100.0%

54.5% 25.0% 34.3%

5 18 23

7.2 15.8 23.0

21.7% 78.3% 100.0%

45.5% 75.0% 65.7%

11 24 35

11.0 24.0 35.0

31.4% 68.6% 100.0%

100.0% 100.0% 100.0%

Count

Expected Count

% within Availabilityin the U.S.

% within Very Goodversus Other

Count

Expected Count



Count

Expected Count



National

Regional


Total

.00 Very Good


Total


Raw Counts


6 6 12

3.8 8.2 12.0

50.0% 50.0% 100.0%

54.5% 25.0% 34.3%

5 18 23

7.2 15.8 23.0

21.7% 78.3% 100.0%

45.5% 75.0% 65.7%

11 24 35

11.0 24.0 35.0

31.4% 68.6% 100.0%

100.0% 100.0% 100.0%

Count

Expected Count



Count

Expected Count



Count

Expected Count



National

Regional


Total

.00 Very Good


Total


Expected Under Independence


6 6 12

3.8 8.2 12.0

50.0% 50.0% 100.0%

54.5% 25.0% 34.3%

5 18 23

7.2 15.8 23.0

21.7% 78.3% 100.0%

45.5% 75.0% 65.7%

11 24 35

11.0 24.0 35.0

31.4% 68.6% 100.0%

100.0% 100.0% 100.0%

Count

Expected Count



Count

Expected Count



Count

Expected Count



National

Regional


Total

.00 Very Good


Total


% Within Rows


6 6 12

3.8 8.2 12.0

50.0% 50.0% 100.0%

54.5% 25.0% 34.3%

5 18 23

7.2 15.8 23.0

21.7% 78.3% 100.0%

45.5% 75.0% 65.7%

11 24 35

11.0 24.0 35.0

31.4% 68.6% 100.0%

100.0% 100.0% 100.0%

Count

Expected Count



Count

Expected Count



Count

Expected Count



National

Regional

Availability

in the U.S.

Total

.00 Very Good


Total


% Within Columns


6 6 12

3.8 8.2 12.0

50.0% 50.0% 100.0%

54.5% 25.0% 34.3%

5 18 23

7.2 15.8 23.0

21.7% 78.3% 100.0%

45.5% 75.0% 65.7%

11 24 35

11.0 24.0 35.0

31.4% 68.6% 100.0%

100.0% 100.0% 100.0%

Count

Expected Count



Count

Expected Count



Count

Expected Count



National

Regional


Total

.00 Very GoodVery Good versus Other

Total



SPSS also allows you to have categorical factors with more than two levels

Rated Quality of Beer * Availability in the U.S. Crosstabulation

Count

6 5 11

4 10 14

2 8 10

12 23 35

VeryGood

Good

Fair

Rated Qualityof Beer

Total

National Regional

Availability in the U.S.

Total



Note warning message

Here you use Likelihood Ratio since there is no Fisher

Chi-Square Tests

3.113a 2 .211

3.086 2 .214

2.750 1 .097

35

Pearson Chi-Square

Likelihood Ratio


N of Valid Cases

Value dfAsymp. Sig.

(2-sided)

3 cells (50.0%) have expected count less than 5. Theminimum expected count is 3.43.

a.


Education and Raises in Construction: do you see any

structure?Number of Promotions: 0, 1, 2, 3+ * Education level Crosstabulation

71 153 224

77.8 146.2 224.0

39 71 110

38.2 71.8 110.0

26 29 55

19.1 35.9 55.0

11 17 28

9.7 18.3 28.0

8 21 29

10.1 18.9 29.0

155 291 446

155.0 291.0 446.0

Count

Expected Count

Count

Expected Count

Count

Expected Count

Count

Expected Count

Count

Expected Count

Count

Expected Count

0 Promotions

1 Promotion

2 Promotions

3 Promotions

4+ Promotions

Number ofPromotions: 0,1, 2, 3+

Total

Less thanBachelor's

Bachelor'sor higher

Education level

Total


Education and Raises in Construction

Chi-Square Tests

5.659a 4 .226

5.533 4 .237

.681 1 .409

446

Pearson Chi-Square

Likelihood Ratio


N of Valid Cases

Value dfAsymp. Sig.

(2-sided)

0 cells (.0%) have expected count less than 5. Theminimum expected count is 9.73.

a.


# Companies & Raises in Construction

Number of Promotions: 0, 1, 2, 3+ * Number of Companies Worked For (Categorical) Crosstabulation

170 40 14 224

179.8 34.7 9.5 224.0

93 16 1 110

88.3 17.0 4.7 110.0

44 8 3 55

44.1 8.5 2.3 55.0

25 2 1 28

22.5 4.3 1.2 28.0

26 3 0 29

23.3 4.5 1.2 29.0

358 69 19 446

358.0 69.0 19.0 446.0

Count

Expected Count

Count

Expected Count

Count

Expected Count

Count

Expected Count

Count

Expected Count

Count

Expected Count

0 Promotions

1 Promotion

2 Promotions

3 Promotions

4+ Promotions


Total

<= 5companies

6-10companies

11+companies

Number of Companies Worked For(Categorical)

Total

Notice how those who have worked for lot of companies have small number of promotions.


Number of Promotions: 0, 1, 2, 3+ * Number of Companies Worked For (Categorical) Crosstabulation

170 40 14 224

179.8 34.7 9.5 224.0

47.5% 58.0% 73.7% 50.2%

93 16 1 110

88.3 17.0 4.7 110.0

26.0% 23.2% 5.3% 24.7%

44 8 3 55

44.1 8.5 2.3 55.0

12.3% 11.6% 15.8% 12.3%

25 2 1 28

22.5 4.3 1.2 28.0

7.0% 2.9% 5.3% 6.3%

26 3 0 29

23.3 4.5 1.2 29.0

7.3% 4.3% .0% 6.5%

358 69 19 446

358.0 69.0 19.0 446.0

100.0% 100.0% 100.0% 100.0%

Count

Expected Count

% within Number ofCompanies WorkedFor (Categorical)

Count

Expected Count


Count

Expected Count


Count

Expected Count


Count

Expected Count


Count

Expected Count


0 Promotions

1 Promotion

2 Promotions

3 Promotions

4+ Promotions


Total

<= 5companies

6-10companies

11+companies

Number of Companies Worked For(Categorical)

Total47.5% of those with <= 5 companies have zero promotions,14.3% have 3 or more

73.7% of those with 11+ companies have zero promotions, 5.3% have 3 or more

May suggest a trend



Chi-Square Tests

11.515a 16 .777

14.989 16 .525

5.460 1 .019

446

Pearson Chi-Square

Likelihood Ratio


N of Valid Cases

Value dfAsymp. Sig.

(2-sided)

16 cells (64.0%) have expected count less than 5. Theminimum expected count is .06.

a.

General chisquared test is not significant

Note the significant “Linear-by-Linear Association”.What is this?



Chi-Square Tests

11.515a 16 .777

14.989 16 .525

5.460 1 .019

446

Pearson Chi-Square

Likelihood Ratio


N of Valid Cases

Value dfAsymp. Sig.

(2-sided)

16 cells (64.0%) have expected count less than 5. Theminimum expected count is .06.

a.

Linear-by-Linear Association (Crosstabs)A measure of linear association between the row and column variables I This statistic should not be used for nominal (unordered) data.

Also known as the Mantel-Haenszel chi-square test. This makes sense: there appears to be some ordered inverse relationship between # of promotions and # of companies



0 5 10 15 20

Number of companies worked for

0

10

20

30

40

50

Construction DataNote the negative trend

Copyright (c) Bani K. Mallick1 STAT 651 Lecture #17.

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Transcript of Copyright (c) Bani K. Mallick1 STAT 651 Lecture #17.